Professor Dave here, let’s talk about integration.
Over the past dozen or so tutorials, we have learned all about differentiation.
What it is, how to find derivatives of all kinds of functions, and some applications.
Now that we’ve wrapped up a survey of differential calculus, it’s time to tackle another beast,
which we call integral calculus.
Mathematicians have been playing with the concept of integration for ages, all the way
back to the Greeks, but it wasn’t until Newton’s time that it was realized that
integration and differentiation are inverse operations.
Just like addition and subtraction, or multiplication and division, integration and differentiation
are linked, and this realization was what unified centuries of tinkering to yield modern calculus.
So let’s start by learning exactly what integration is.
In geometry we learned about the concept of area, which is the amount of two-dimensional
surface covered by a figure.
We also learned how to calculate the area of all kinds of different polygons, using
specific formulas.
But the ancient mathematicians realized that things get much trickier when curvature is involved.
Say we have this curve here, given by the function f of x, and we want to know the area
under this curve over this interval here, from a to b.
That means we are looking at this region, S, that is enclosed by the curve, the x-axis,
and the vertical lines x equals a and x equals b.
We quickly realize that there is no formula for this, as this is not a rectangle, or a
trapezoid, or any other polygon, as polygons have only straight line segments for sides.
But when we first learned about differentiation, we saw that we could get the slope of a tangent
line through a method of exhaustion.
Can there be some parallel method here?
In fact there is.
When we couldn’t get the slope of a line with one point, we made a second point so
that we could get the slope, and then pushed the second point towards the first.
Here, we can’t get the area of this shape, so let’s approximate it with a shape we
can get the area of, a rectangle.
Let’s place some rectangles, and we might see that it’s not the best approximation.
But let’s put more of them and make them narrower.
That is starting to look better, as they don’t stick out from the curve so much.
We see that as they get more and more narrow, we are more closely approximating this area.
In the limit of an infinite number of infinitely thin rectangles, we will get the precise area
under this curve.
Let’s do this more quantitatively on a specific function to get a better grasp of what’s
going on.
Let’s take y equals x squared.
Say we want to know the area under this curve over the interval of zero to one, so let’s
write in the point (one, one), and label this shaded region S. We know that this area must
be somewhere in between zero and one, because a square with these sides will have an area
of one, and this area is smaller than that.
But let’s use some rectangles and see how close we can get.
First let’s chop this area up into four sections, from zero to one fourth, then to
one half, then to three fourths, then to one.
We can approximate each of these with a rectangle, which is useful, because it will be easy to
get the areas of these rectangles.
Each of them has a base of one fourth, and we can get the heights by using the function.
The x coordinate of this first point is one fourth, and every point on this curve has
the coordinates x, x squared, so the height of this first rectangle must be one fourth
squared, or one sixteenth.
The second rectangle has a height of one half squared, or one fourth.
The third rectangle has a height of nine sixteenths, and the fourth has a height of one.
We know that the area of a rectangle is base times height, so we just multiply these values
together to get the area of each rectangle, and add them up to get the total area.
This is around 0.469.
So this is our first rough approximation.
We know that the true value is less than this, because these rectangles stick out above the curve.
But now let’s use ten rectangles, and see what that gives us.
These rectangles will all have a base of one tenth, as well as the following heights, and
therefore these corresponding areas.
Again, we simply add up the areas, and we get 0.385.
This looks like it is much closer to the area we are looking for, because the rectangles
don’t stick out past the curve as much as before.
We can use smaller and smaller rectangles to get closer and closer to the precise area,
and while we won’t show hundreds of rectangles here, this table illustrates what happens
as n, the number of rectangles, gets larger, all the way up to a thousand.
We can clearly see that this area is getting closer and closer to a specific value, and
that value is one third.
In the limit of an infinite number of rectangles, the sum of the areas of the rectangles will
be precisely one third, therefore the area under this curve is equal to one third.
We can apply this generally to any function, like the first one we looked at.
However many rectangles we use, we can represent the sum of their areas using summation notation,
which as we recall from an earlier tutorial, involves this upper case sigma, which just
means to add up the series of terms that will follow.
Here we are adding up areas, so we need a term to represent the height of a rectangle,
which is given by f of xi, and the base of a rectangle, which is given by delta x.
These multiply, and we add up the sum of these terms from i equals one, to n, and in the
limit of n equals infinity, we have our area under the curve.
And that’s our basic introduction to integration.
It might seem like a departure from what we’ve been talking about, as differentiation always
had to do with rates of change.
What does the area under a curve have to do with the rate of change of that curve?
In fact, there is a deep connection here, and the elegant articulation of that connection
by Isaac Newton is what solidifed the status of calculus as a systematic mathematical method.
Let’s move forward and see exactly what this connection looks like.